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alex41 [277]
3 years ago
9

The doppler effect is when objects are not moving? True False

Physics
1 answer:
SVETLANKA909090 [29]3 years ago
6 0

Answer:

reviewing the statement we can say that it is FALSE

Explanation:

The Doppler effect is the change in the perceived frequency of a sound wave by the RELATIVE MOTION of the source and the observer.

This effect is very useful for finding the speed of an object.

When reviewing the statement we can say that it is FALSE

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A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl
zysi [14]

Answer:\alpha =9.302\ rad/s^2

Explanation:

Given

mass of sphere m=250\ gm

diameter of sphere d=4.30\ cm

radius r=\frac{4.30}{2}\ cm

f=0.0200\ N

friction will provide resisting torque so

f\times r=I\times \alpha

where I=\text{moment of Inertia}

f=\text{friction force}

\alpha =\text{angular acceleration}

I=\frac{2}{5}mr^2

0.02\times r=\frac{2}{5}mr^2\times \alpha

\alpha =\frac{5}{2r}\times f

\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02

\alpha =9.302\ rad/s^2

(b)time taken to decrease its rotational speed by 21\ rad/s

t=\dfrac{\Delta \omega }{\alpha }

t=\dfrac{21}{9.302}

t=2.25\ s

6 0
3 years ago
What kind of image is formed by a plane mirror
lisov135 [29]
A distorted image. :3

hope i help
5 0
4 years ago
What is precipitation reactions​
Virty [35]
<h2>precipitation reactions</h2>

α prєcípítαtíσn rєαctíσn rєfєrs tσ thє fσrmαtíσn σf αn ínsσluвlє sαlt whєn twσ sσlutíσns cσntαíníng sσluвlє sαlts αrє cσmвínєd. thє ínsσluвlє sαlt thαt fαlls σut σf sσlutíσn ís knσwn αs thє prєcípítαtє, hєncє thє rєαctíσn's nαmє. prєcípítαtíσn rєαctíσns cαn hєlp dєtєrmínє thє prєsєncє σf vαríσus íσns ín sσlutíσn.

4 0
4 years ago
Read 2 more answers
A 1,550 kg car moving south at 10.0 m/s collides with a 2,550 kg car moving north. The cars stick together and move as a unit af
nikitadnepr [17]

Answer:

14.47 m/s

Explanation:

The momentum must be preserved before and after the collision:

The total momentum before the collision

m_1v_1 + m_2v_2

where m_1 = 1550 kg, m_2 = 2550 kg are the masses of the car moving south and north, respectively, before the collision. v_1 = -10m/s is the velocity of the car moving South. We take the velocity to the North as the positive direction

The total momentum after the collision

(m_1 + m_2)V

where V = 5.22m/s is the velocity of both cars after the collision

We can equalize the 2 equations and plug in the numbers:

m_1v_1 + m_2v_2 = (m_1 + m_2)V

1550(-10) + 2550*v_2 = (1550 + 2550)5.22

-15500 + 2550v_2 = 21402

2550v_2 = 36902

v_2 = 36902/2550 = 14.47 m/s

6 0
3 years ago
A 250 g toy car is placed on a narrow 60-cm-diameter track with wheel grooves that keep the car going in a circle. The 1.5 kg tr
Lady_Fox [76]

Answer:

\omega_{t}=-3.9 rpm

Explanation:

Let's use the conservation of angular momentum here. If there is a conservation the addition of both must be zero.

L_{c}+L_{t}=0

The momentum of inertia of both are:

I_{c}=m_{c}R^{2}

I_{t}=m_{t}R^{2}

The angular momentum is the product between angular velocity and momentum of inertia.

I_{c}\omega_{c}+I_{t}\omega_{t}=0

I_{c}\omega_{c}+I_{t}\omega_{t}=0

Let's solve it for ω(t).

\omega_{t}=-\frac{I_{c}\omega_{c}}{I_{t}}

\omega_{t}=-\frac{m_{c}R^{2}\omega_{c}}{m_{t}R^{2}}

But \omega=V/R

  • R is the radius R=D/2 = 30 cm = 0.3 m
  • V is the steady speed V = 0.74 m/s

\omega_{t}=-\frac{m_{c}R^{2}*V/R}{m_{t}R^{2}}

\omega_{t}=-\frac{m_{c}V}{m_{t}R}

\omega_{t}=-\frac{0.25*0.74}{1.5*0.3}  

\omega_{t}=-0.41 rad/s

 

Knowing that 2π  rad is a rev. We have.

\omega_{t}=-3.9 rpm

The minus sign means the track is moving opposite of the car.      

I hope it helps you!

4 0
4 years ago
Read 2 more answers
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